Question

Derive the relationship between relative lowering of vapour pressure and molar mass of non-volatile solute.

Answer 1

Consider a solution in which $W_{2}g$ of solute of molar mass $M_2$ is dissolved in $W_{1}g$ of solvent of molar mass $M_1$.

Hence, number of moles of solvent, $n_1=\frac{W_1}{M_1}$ and,

Number of moles of solute, $n_2=\frac{W_2}{M_2}$

$\therefore$ Total number of moles $n=n_1+n_2$

Mole fraction of the solvent, $x_1=\frac{n_1}{n}$

Mole fraction of the solute, $x_2=\frac{n_2}{n}$

In case of a dilute solution, the concentration of number of moles of the solute is very low, i.e., $n_1>>n_2$

$\therefore n_1+n_2\simeq n_1$

Now, the mole fraction of the solute $x_2$ is given by,

$x_2=\frac{n_2}{n_1+n_2}\simeq \frac{n_2}{n_1}$

$\therefore x_2=\frac{W_2/M_2}{W_1/M_1}$

$\therefore x_2=\frac{W_2\times M_1}{W_1\times M_2}$

If $P_1^0$ and P are the vapour pressure of pure solvent and a solution respectively, then relative lowering of vapour pressure is given by,

$P=\frac{P_1^0-P}{P_1^0}$

By Raoult's law $\frac{\Delta P}{P_1^0}=\frac{P_1^0-P}{P_1^0}=x_2$

$\frac{P_1^0-P}{P_1^0}=\frac{W_2{M}_1}{W_1{M}_2}$

or $\therefore \frac{\Delta P}{P_1^0}=\frac{n_2}{n_1}=\frac{W_2/M_2}{W_1/M_1}=\frac{W_2{M}_1}{W_1{M}_2}$

Hence, by determining the vapour pressure of a pure solvent and a solution, the molar mass of a non-volatile solute can be measured.